Question

Use spherical coordinates.

Evaluate (S = integral sign)
SSS(x2 + y2) dV

where E lies between the spheres
x2 + y2 + z2 = 1 and x2 + y2 + z2 = 25.

The answer i came up with was 624.8pi^2

Answer 1

ok so what have u done so far? u can show your work

Answer 2

start by converting them into spherical coordinates

Answer 3

\[x=r.\sin \theta. \cos \phi\\y=r.\sin \theta.\sin \phi\\z=r.\cos \theta\]

Answer 4

Let me take a picture :)

Answer 5

yes the same thing @Concentrationalizing :)

Answer 6

Just looked odd how ya had it, lol.

Answer 7

lol :)

Answer 8

try to find the limits first

Answer 9

\[x = \rho \sin \phi \cos \theta\]
\[y = \rho \sin \phi \sin \theta\]
\[z = \rho \cos \phi \]


had to correct myself :P

Answer 10

okey :)

Answer 11

\[\int\limits_{0}^{\pi/2}\int\limits_{0}^{\pi/2}\int\limits_{1}^{5}\]
seems ok?

Answer 12

Sorry my picture is REALLY blurry and can't see. I got the limites of integration down.
I believe I foudn my mistake give me a second

when I tookt he integration for sin^2(phi) i accentally wrote phi as theta..

Answer 13

May I ask why you limited your phi and theta limits to only pi/2?

Answer 14

i think i messed up lol
it should be 0 to pi ,and 0 to 2 pi
right?

Answer 15

didnt' see that my limits are

1 <= rho <=5
o <= theta <= 2pi
o <= phi <=pi

Answer 16

Yes :)

Answer 17

Those are the correct limits :3

Answer 18

yeah

Answer 19

now easy integration

Answer 20

Right so far?

Answer 21

Same answer...

Answer 22

Hmm... \(x^{2} + y^{2} = r^{2}\)
\( r= \rho sin \phi\ \implies\ r^{2} = \rho^{2} sin^{2} \phi \implies x^{2} + y^{2} = \rho^{2} sin^{2} \phi \)
\[\int\limits_{0}^{2 \pi}\int\limits_{0}^{\pi}\int\limits_{1}^{5} \rho ^{2} \sin^{2} \phi (\rho ^{2} \sin \phi ) d \rho d \phi d \theta\]

The \(\rho ^{2} sin ^{2} \phi\) comes from the conversion of the original function \(x^{2} + y^{2} \). The rest comes from the necessary part of an integration using spherical coordinates.

Answer 23

So I actually messed up on my setup jumped to quick

Answer 24

yes did some wrong calculations

Answer 25

Okay let me do it again don't give me the answer yet :X

Answer 26

Just combine and make it into \(\rho ^{4} sin^{3} \phi\) and integrate through :)

Answer 27

from where P^4 is coming

Answer 28

Integration in spherical coordinates requires we have \(\rho ^{2} sin \phi\ d \rho\ d \phi\ d \theta\). The conversion of the function inside of the integral gives us: \(x^{2} + y^{2} = p^{2} sin^{2} \phi\) Both of these are being multiplied inside of the integral, giving:
\[\int\limits_{0}^{2 \pi}\int\limits_{0}^{\pi} \int\limits_{1}^{5}\rho ^{2} \sin^{2} \phi (p^{2} \sin \phi d \rho d \phi d \theta) = \int\limits_{0}^{2 \pi}\int\limits_{0}^{\pi}\int\limits_{1}^{5} \rho ^{4} \sin^{3} \phi d \rho d \phi d \theta \]

Answer 29

yes okayyyy :)

Answer 30

\[(3124\pi)/5\int\limits_{0}^{2\pi}\sin^3(\phi)dphi\]

Answer 31

I'm not sure how to handle the \[\sin^3\phi\]

Answer 32

We can do the half angle identity and get

Answer 33

\[(sinphi-sinphicos2\phi)/2\]

Answer 34

Then do a integration by parts?

Answer 35

\[\sin^{3} \phi = \sin \phi (1 - \cos^{2} \phi) = \sin \phi - \sin \phi \cos^{2} \phi\]

Doing this, you can do two integrations now, the integral of \(sin \phi\) alone and a u-substitution in the seciond integral, \(u = cos \phi\)

Answer 36

\[\sin^3 \phi=\frac{ 3\sin \phi-\sin3 \phi }{ 4 }\]

Answer 37

this will ease ur integration :)

Answer 38

I don't think I recognize that conversion, can you explain please? It's beautiful btw :')

Answer 39

Pfft, fancy identities we don't learn, haha. I'll just stick with what I know, lol.

Answer 40

lol I tried the integration by parts but it just gets me to a loop.

Answer 41

its a basic trig indentity
\[\sin3A=\sin(2A+A)=\sin2A.cosA+\cos2A.sinA\\=2.sinA.cosA.cosA+(1-2\sin^2A).sinA\\=2sinA(1-\sin^2A)+sinA-2\sin^3A\\=3sinA-4\sin^3A\\sin3A=3sinA-4\sin^3A\\sin^3A=\frac{ 3sinA-\sin3A }{ 4 }\]
proved it :)

Answer 42

But the u sub work s a LOT better..

Answer 43

yes its hard to remember them :)

Answer 44

Yeah, the splitting apart of sin^3 into sin and (1-cos^2) works well. And I remember pretty much any identity that ive learned, but the sin^3 one I havent seen as a basic identity.

Answer 45

Its always the usual ones, double angle, half angle, power reducing, etc, etc.

Answer 46

yes its a important one just makes your calculation easy :)

Answer 47

So if I were to say \[\sin^4\phi\] = \[(4sinphi-sin4phi)/5\]

Answer 48

You should end up with a \(sin^{4} \phi\). At least not that I see.

Answer 49

*shouldnt

Answer 50

lol u have to derive it then for sin^n(phi)

Answer 51

I got the answer is zero.. what did I do wrong -.-

Answer 52

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{1}^{5}\rho ^{4} \sin^{3} \phi d \rho d \phi d \theta \] The integration with respect to rho gives
\[\frac{ 5^{5} }{ 5 }\sin^{3} \phi - \frac{ 1 }{ 5 }\sin^{3} \phi = \frac{ 3124 }{ 5 }\sin^{3} \phi\]

Then we get:
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi} \frac{ 3124 }{ 5 }\sin^{3} \phi d \phi d \theta = \int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\frac{ 3124 }{ 5 }\sin \phi (1- \cos^{2} \phi) d \phi d \theta\]
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\frac{ 3124 }{ 5 }(\sin \phi - \sin \phi \cos^{2} \phi) d \phi d \theta\] Integrating sin phi is straightforward, integrating the sin cos^2 is just u-sub with u = cos phi
\[\frac{ 3124 }{ 5 }(-\cos \phi + \frac{ \cos^{3} \phi }{ 3 })\]
THen you have limits from 0 to pi for that integral, then you can do the last integraton with respect to theta. Just to get ya goin :)

Answer 53

you've got some concentration @concentrationalizing :P

Answer 54

:) As long as things don't get too "Complexicated" o_o

Answer 55

lol :)

Answer 56

I'm making so many mistakes stupid rho and phi and rho and asdfasdfhkasgf

Answer 57

\[\int\limits_{0}^{2\pi}sinphi = 0\]

Answer 58

yes?

Answer 59

Because cos(2pi) = 1 and cos(0) = 1 and 1-1 =0

Answer 60

xDD No worries :) So with where I left off, we have:

\[\frac{ 3124 }{ 5 }[(-\cos(\pi)+ \frac{ \cos^{3}(\pi) }{ 3 }) - (-\cos (0) + \frac{ \cos^{3}(0) }{ 3 })]\]
\[\frac{ 3124 }{ 5 }[1 - \frac{ 1 }{ 3 }-(-1+ \frac{ 1 }{ 3 })] = \frac{ 3124 }{ 5 }(2 - \frac{ 2 }{ 3 }) = \frac{ 12496 }{ 15 }\] That leavs the final integral to come down to simply
\[\int\limits_{0}^{2\pi}\ \frac{ 12496 }{ 15 }d \theta\]

Answer 61

So 24992pi/15

Answer 62

Sounds good to me :)

Answer 63

Can you follow the integration steps, though?

Answer 64

I figured out my mistake...
Looking at my limits of integration I switch phi and theta... so I had

2pi <= phi <= 0

Answer 65

And yes the steps of integration were perfect thank you, it made it clear to me what my mistake was lol

Answer 66

I was wondering why you had your limits of integration a bit off, then I realize i was actually the one off.

Answer 67

Alright, awesome ^_^ Lol, well I make mistakes as well, but I know I dont have any thetas in my integration. And theta is so often 0 to 2pi, so it kind of registers to me to match 0 to pi with phi and 0 to 2pi with theta. That and for the most part, you pretty restrict phi to be between 0 and pi, so not sure, I do my best to not mix stuff up, lol.

Answer 68

Must not rush these problems again... keeping my limits of integration and what I'm integration with respect is tricky enough lol always feel like I'm being caught in a tangle.

Answer 69

Thanks! The answer was correct

Answer 70

Awesome :P Glad I could help :)

Answer 71

You two are officially my first two fans much appreciated I hope to get some more help you guys later on Exam on Thursday so study study study :)

Answer 72

Lol, yeah, good luck studying. I got exams on Wednesday o.o So we all study study cram cram now :D